Show $\displaystyle \sum_{n=1}^\infty \frac{nx^2}{n^3+x^3}$ is uniformly convergent on $[0,A]$ for any $A \gt 0$.
Then prove
$$ \lim_{x \to 1} \sum_{n=1}^\infty \frac{nx^2}{n^3+x^3} = \sum_{n=1}^\infty \frac{n}{n^3+1}.$$
My attempt:
Let $A \gt 0$ be given. Let $\displaystyle M_n := \frac{A^2}{n^2}$.
Then $\displaystyle \left| \frac{nx^2}{n^3+x^3} \right| \lt \left| \frac{nx^2}{n^3} \right|=\frac{x^2}{n^2} \leq M_n, \forall x \in [0,A]$. Also $\sum_{n=1}^\infty M_n < \infty$, so by Weierstrass M test, the series converges uniformly on $[0,A]$.
Now for the second part, to simplify things, I wrote it down like this:
Let $f(x) = \sum_{n=1}^\infty \frac{nx^2}{n^3+x^3}$. Prove $\lim_{x \to 1} f(x) = f(1)$.
Now isn't this equivalent of asking if $f(x)$ is continuous at $x=1$?
We know $f(x)$ is uniformly convergent on $[0,A]$ so it is continuous on $[0,A]$. Pick $A=2$, then $f(x)$ is continuous at $x=1$, and hence the limit converges to $f(1)$.
Is there anything wrong with my second proof/argument?